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A good day to everyone.

Consider the following "Conjecture":

If $a, b \in M \subset \mathbb{N}$, then $1 < a < b$ and ... [plus some more conditions on $a, b$ and $M$...] if and only if $\sigma(a) < \sigma(b)$.

I have two questions at this point:

(1) What properties should elements of the set $M$ have to satisfy this "Conjecture"?

(2) What happens to the sets in $M$ if you restrict to the case $\gcd(a, b) = 1$? (Of course, the sets in (1) should have greater asymptotic density than the sets in (2), but can this notion be made more precise?)

Now, for the motivation: An example of a set $M$ satisfying this "Conjecture" is any pair $(A, B)$ of relatively prime factors of an odd perfect number given in the Eulerian form $N = {q^k}{n^2}$ (modulo some exceptions), where $\gcd(q, n) = 1$ and $q \equiv k \equiv 1 \pmod4$.

Let $$I(x) = \displaystyle\frac{\sigma(x)}{x}$$ be the abundancy index of $x \in \mathbb{N}$. Then by enumerating all possible permutations of the set

$$q^k, n, \sigma(q^k), \sigma(n) \in \mathbb{N}$$

we know that exactly one of the following holds (because of $1 < I(q^k) < I(n)$):

$$[1] \hspace{0.2in} q^k < n < \sigma(q^k) < \sigma(n)$$ $$[2] \hspace{0.2in} q^k < \sigma(q^k) < n < \sigma(n)$$ $$[3] \hspace{0.2in} n < q^k < \sigma(q^k) < \sigma(n)$$ $$[4] \hspace{0.2in} n < q^k < \sigma(n) < \sigma(q^k)$$ $$[5] \hspace{0.2in} n < \sigma(n) < q^k < \sigma(q^k)$$

Last question: Which of these five "configurations" could we eliminate?

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Arnie, where did you pick up the notation I(n) for the abundancy index? I haven't seen that used before. –  Charles Mar 7 '11 at 0:10
    
@Charles, as far as I know Richard Ryan and Judy Holdener uses that notation in their papers. For a (comprehensive?) list of resources on the abundancy index, you can refer to Walter Nissen's page at upforthecount.com/math/abundance.html. –  Jose Arnaldo Dris Mar 7 '11 at 1:35
1  
Thanks! I've read most of the papers linked there, including Holdner's "A theorem of Touchard on the form of odd perfect numbers", but not the Ryan paper. –  Charles Mar 7 '11 at 2:45
    
You're most welcome =) –  Jose Arnaldo Dris Mar 7 '11 at 3:40
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2 Answers

up vote 3 down vote accepted

I think this is a ludicrously broad question. For (1), you could take $M$ to be the set of "champions" for the divisor function, $M=\lbrace1,2,3,4,6,8,10,12,16,18,20,24,\dots\rbrace$, which is http://oeis.org/A002093 - no conditions on $a$, $b$ necessary. At the other extreme, you could take $M$ to be the whole of the positive integers, with the condition $a\mid b$. There may be some interesting answers in between, but this strikes me as more of a fishing expedition than a serious question.

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Thanks Gerry! Rest assured that it IS a serious question. =) –  Jose Arnaldo Dris Mar 5 '11 at 23:53
    
+1 for the first and last sentences. –  Yemon Choi Mar 6 '11 at 3:39
    
Thanks Yemon! @grin@ –  Jose Arnaldo Dris Jun 12 '11 at 21:38
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Let $N = {q^k}{n^2}$ be an odd perfect number (i.e. OPN) given in the so-called Eulerian form (i.e. $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q, n) = 1$).

Note that $k = 1$ rules out [1] and [4], because $q$ and $\sigma(q) = q + 1$ are consecutive integers. I am guessing a possible approach will be to derive [a] condition(s) equivalent to $k = 1$ and work from there.

A sufficient condition for $k = 1$ is $n < q$ (or $n < q^2$). I will be most surprised if somebody would be able to establish that either of these two is also necessary for $k = 1$.

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